Solve:
$\displaystyle 2y'y''=1+(y')^2$ by letting $\displaystyle y'=v$
Does $\displaystyle y''=v'$
So does $\displaystyle 2vv'=1+v^2$
Not sure I would use the division symbol there. On the LHS, you have $\displaystyle dv/dx$, originally. When you multiply both sides of
$\displaystyle \displaystyle{\frac{2v}{1+v^{2}}\,\frac{dv}{dx}=1}$ by $\displaystyle dx$, what are you going to get?
[EDIT]: acevipa edited post # 5. It's correct now.